64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -0.149 999 999 6 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -0.149 999 999 6(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.149 999 999 6| = 0.149 999 999 6

2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.149 999 999 6.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.149 999 999 6 × 2 = 0 + 0.299 999 999 2;
  • 2) 0.299 999 999 2 × 2 = 0 + 0.599 999 998 4;
  • 3) 0.599 999 998 4 × 2 = 1 + 0.199 999 996 8;
  • 4) 0.199 999 996 8 × 2 = 0 + 0.399 999 993 6;
  • 5) 0.399 999 993 6 × 2 = 0 + 0.799 999 987 2;
  • 6) 0.799 999 987 2 × 2 = 1 + 0.599 999 974 4;
  • 7) 0.599 999 974 4 × 2 = 1 + 0.199 999 948 8;
  • 8) 0.199 999 948 8 × 2 = 0 + 0.399 999 897 6;
  • 9) 0.399 999 897 6 × 2 = 0 + 0.799 999 795 2;
  • 10) 0.799 999 795 2 × 2 = 1 + 0.599 999 590 4;
  • 11) 0.599 999 590 4 × 2 = 1 + 0.199 999 180 8;
  • 12) 0.199 999 180 8 × 2 = 0 + 0.399 998 361 6;
  • 13) 0.399 998 361 6 × 2 = 0 + 0.799 996 723 2;
  • 14) 0.799 996 723 2 × 2 = 1 + 0.599 993 446 4;
  • 15) 0.599 993 446 4 × 2 = 1 + 0.199 986 892 8;
  • 16) 0.199 986 892 8 × 2 = 0 + 0.399 973 785 6;
  • 17) 0.399 973 785 6 × 2 = 0 + 0.799 947 571 2;
  • 18) 0.799 947 571 2 × 2 = 1 + 0.599 895 142 4;
  • 19) 0.599 895 142 4 × 2 = 1 + 0.199 790 284 8;
  • 20) 0.199 790 284 8 × 2 = 0 + 0.399 580 569 6;
  • 21) 0.399 580 569 6 × 2 = 0 + 0.799 161 139 2;
  • 22) 0.799 161 139 2 × 2 = 1 + 0.598 322 278 4;
  • 23) 0.598 322 278 4 × 2 = 1 + 0.196 644 556 8;
  • 24) 0.196 644 556 8 × 2 = 0 + 0.393 289 113 6;
  • 25) 0.393 289 113 6 × 2 = 0 + 0.786 578 227 2;
  • 26) 0.786 578 227 2 × 2 = 1 + 0.573 156 454 4;
  • 27) 0.573 156 454 4 × 2 = 1 + 0.146 312 908 8;
  • 28) 0.146 312 908 8 × 2 = 0 + 0.292 625 817 6;
  • 29) 0.292 625 817 6 × 2 = 0 + 0.585 251 635 2;
  • 30) 0.585 251 635 2 × 2 = 1 + 0.170 503 270 4;
  • 31) 0.170 503 270 4 × 2 = 0 + 0.341 006 540 8;
  • 32) 0.341 006 540 8 × 2 = 0 + 0.682 013 081 6;
  • 33) 0.682 013 081 6 × 2 = 1 + 0.364 026 163 2;
  • 34) 0.364 026 163 2 × 2 = 0 + 0.728 052 326 4;
  • 35) 0.728 052 326 4 × 2 = 1 + 0.456 104 652 8;
  • 36) 0.456 104 652 8 × 2 = 0 + 0.912 209 305 6;
  • 37) 0.912 209 305 6 × 2 = 1 + 0.824 418 611 2;
  • 38) 0.824 418 611 2 × 2 = 1 + 0.648 837 222 4;
  • 39) 0.648 837 222 4 × 2 = 1 + 0.297 674 444 8;
  • 40) 0.297 674 444 8 × 2 = 0 + 0.595 348 889 6;
  • 41) 0.595 348 889 6 × 2 = 1 + 0.190 697 779 2;
  • 42) 0.190 697 779 2 × 2 = 0 + 0.381 395 558 4;
  • 43) 0.381 395 558 4 × 2 = 0 + 0.762 791 116 8;
  • 44) 0.762 791 116 8 × 2 = 1 + 0.525 582 233 6;
  • 45) 0.525 582 233 6 × 2 = 1 + 0.051 164 467 2;
  • 46) 0.051 164 467 2 × 2 = 0 + 0.102 328 934 4;
  • 47) 0.102 328 934 4 × 2 = 0 + 0.204 657 868 8;
  • 48) 0.204 657 868 8 × 2 = 0 + 0.409 315 737 6;
  • 49) 0.409 315 737 6 × 2 = 0 + 0.818 631 475 2;
  • 50) 0.818 631 475 2 × 2 = 1 + 0.637 262 950 4;
  • 51) 0.637 262 950 4 × 2 = 1 + 0.274 525 900 8;
  • 52) 0.274 525 900 8 × 2 = 0 + 0.549 051 801 6;
  • 53) 0.549 051 801 6 × 2 = 1 + 0.098 103 603 2;
  • 54) 0.098 103 603 2 × 2 = 0 + 0.196 207 206 4;
  • 55) 0.196 207 206 4 × 2 = 0 + 0.392 414 412 8;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.149 999 999 6(10) =

0.0010 0110 0110 0110 0110 0110 0110 0100 1010 1110 1001 1000 0110 100(2)


6. Positive number before normalization:

0.149 999 999 6(10) =

0.0010 0110 0110 0110 0110 0110 0110 0100 1010 1110 1001 1000 0110 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.149 999 999 6(10) =

0.0010 0110 0110 0110 0110 0110 0110 0100 1010 1110 1001 1000 0110 100(2) =

0.0010 0110 0110 0110 0110 0110 0110 0100 1010 1110 1001 1000 0110 100(2) × 20 =

1.0011 0011 0011 0011 0011 0011 0010 0101 0111 0100 1100 0011 0100(2) × 2-3


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -3

Mantissa (not normalized):
1.0011 0011 0011 0011 0011 0011 0010 0101 0111 0100 1100 0011 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1020(10) =

011 1111 1100(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0011 0011 0011 0011 0011 0011 0010 0101 0111 0100 1100 0011 0100 =

0011 0011 0011 0011 0011 0011 0010 0101 0111 0100 1100 0011 0100


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1100

Mantissa (52 bits) =
0011 0011 0011 0011 0011 0011 0010 0101 0111 0100 1100 0011 0100


The base ten decimal number -0.149 999 999 6 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 011 1111 1100 - 0011 0011 0011 0011 0011 0011 0010 0101 0111 0100 1100 0011 0100

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number -0.149 999 999 7 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number -0.149 999 999 5 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

Number 299 976 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:40 UTC (GMT)
Number -635.8 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:40 UTC (GMT)
Number 53.384 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:40 UTC (GMT)
Number 32 959 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:40 UTC (GMT)
Number 0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 009 874 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:40 UTC (GMT)
Number 279 641 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:39 UTC (GMT)
Number 0.718 281 828 456 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:39 UTC (GMT)
Number -7 067 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:39 UTC (GMT)
Number 9 932 178 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:39 UTC (GMT)
Number 6.9 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 18 19:39 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100